Question

4. A body is thrown vertically up to a height of 250 m in 5 s and then comes back at the same point where from it was thrown. Find
a. the total distance
b. the total displacement
c. the average speed
d. the average velocity

Answer 1

Given that: A body is thrown vertically up to a height of 250 m in 5 s and then comes back at the same point where from it was thrown.

According to the statement:

• Height = 250 metres
• Time = 5 seconds

To calculate:

• The total distance
• The total displacement
• The average speed
• The average velocity

Solution:

• The total distance = 500 m

• The total displacement = 0 m

• The average speed = 100 m/s

• The average velocity = 0 m/s

Using concepts:

• Total distance formula
• Displacement concept
• Average speed formula
• Average velocity formula

Using formulas:

• ${\small{\underline{\boxed{\sf{\pmb{Total \: distance \: = Distance \: 1^{st} \: + Distance \: 2^{nd}}}}}}}$

• ${\small{\underline{\boxed{\sf{\pmb{v \: = \dfrac{s}{t}}}}}}}$

Where, v denotes average speed, s denotes total distance and t denotes total time taken.

• ${\small{\underline{\boxed{\sf{\pmb{v_{av} \: = \dfrac{s}{t}}}}}}}$

Where, v_{av} denotes average velocity, s denotes displacement and t denotes time taken.

• Displacement is the shortest distance between initial point and the final point.

Required solution:

~ Firstly let us find out the total distance!

Explanation: As it is given that the body is vertically up to a height of 250 m and then comes back at the same point where from it was thrown, it covers the same distance again. Henceforth, the total distance becames

$:\implies \sf Total \: distance \: = Distance \: 1^{st} \: + Distance \: 2^{nd} \\ \\ :\implies \sf Total \: distance \: = 250 + 250 \\ \\ :\implies \sf Total \: distance \: = 500 \: metres \\ \\ {\pmb{\sf{Henceforth, \: solved!}}}$

~ Now let's calculate the displacement!

Explanation: As we know that displacement is the shortest distance between initial point and the final point and here body is thrown vertically up then comes back at the same point where from it was thrown. Means the initial and the final positions are same therefore,

$:\implies \sf Total \: displacement \: = zero \\ \\ :\implies \sf Total \: displacement \: = 0 \: metre \\ \\ {\pmb{\sf{Henceforth, \: solved!}}}$

~ Now let's find out the average speed!

Explanation: To solve this part of this question we have to use formula to find out the average speed!

$:\implies \sf v \: = \dfrac{s}{t} \\ \\ :\implies \sf v \: = \dfrac{500}{5} \\ \\ :\implies \sf v \: = \cancel{\dfrac{500}{5}} \: (Cancelling) \\ \\ :\implies \sf v \: = 100 \: ms^{-1} \\ \\ :\implies \sf Average \: speed \: = 100 \: ms^{-1} \\ \\ {\pmb{\sf{Henceforth, \: solved!}}}$

~ Now let's find out the average velocity!

Explanation: To solve this part of this question we have to use formula to find out the average velocity!

$:\implies \sf v_{av} \: = \dfrac{s}{t} \\ \\ :\implies \sf v_{av} \: = \dfrac{0}{5} \\ \\ :\implies \sf v_{av} \: = 0 \: ms^{-1} \\ \\ :\implies \sf Average \: velocity \: = 0 \: ms^{-1} \\ \\ {\pmb{\sf{Henceforth, \: solved!}}}$